To evaluate the limit of

$\lim_{x \to c} \frac{6x^{2} - 7x - 10}{x^{2} + 5x - 14},$

we first analyze the function.

1. Factorization

Both numerator and denominator are quadratic expressions, and factoring them can simplify the expression.

Numerator: $6x^{2} - 7x - 10$

We look for two numbers whose product is $6 \times -10 = -60$ and sum is $-7$. These numbers are $-12$ and $5$.

$6x^{2} - 7x - 10 = 6x^{2} - 12x + 5x - 10$ $= 6x(x - 2) + 5(x - 2)$ $= (6x + 5)(x - 2).$

Denominator: $x^{2} + 5x - 14$

We look for two numbers whose product is $-14$ and sum is $5$. These numbers are $7$ and $-2$.

$x^{2} + 5x - 14 = (x + 7)(x - 2).$

2. Simplified Expression

$\frac{6x^{2} - 7x - 10}{x^{2} + 5x - 14} = \frac{(6x + 5)(x - 2)}{(x + 7)(x - 2)}.$

Cancel the common factor $x - 2$ (valid unless $x = 2$): $\frac{6x + 5}{x + 7}.$

3. Evaluate the Limit

Now, depending on the value of $x \to c$, the limit changes:

Case: $x \to 2$

If $x = 2$: $\lim_{x \to 2} \frac{6x + 5}{x + 7} = \frac{6(2) + 5}{2 + 7} = \frac{12 + 5}{9} = \frac{17}{9}.$

Case: $x \to \infty$ or $x \to -\infty$

If $x \to \infty$ or $x \to -\infty$, divide numerator and denominator by $x$: $\lim_{x \to \infty} \frac{6x + 5}{x + 7} = \lim_{x \to \infty} \frac{6 + \frac{5}{x}}{1 + \frac{7}{x}} = \frac{6}{1} = 6.$

Case: $x \to -\infty$

Similarly: $\lim_{x \to -\infty} \frac{6x + 5}{x + 7} = \frac{6}{1} = 6.$

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Final Answer:

• For $x \to 2$: $\frac{17}{9}$,
• For $x \to \infty$ or $x \to -\infty$: $6$.

Would you like further explanation on these steps or additional cases?

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Related Questions:

1. What happens when $x$ approaches a removable discontinuity like $x = 2$?
2. How do we calculate limits at infinity for rational functions?
3. Can this method be applied to higher-order polynomials?
4. What if the denominator had a higher degree than the numerator?
5. How does the behavior of the function change as $x$ approaches $-\infty$?

Tip: Always check for common factors in rational expressions before evaluating limits.